Solve the following pair of linear equations by the substitution method.
$3 x - y = 3$
$9 x - 3 y = 9$

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Solution

$3 x - y = 3$ ………… $(1)$
$9 x - 3 y = 9$ ……….. $(2)$
on comparing the above equations with standard equation $a x + b y = c = 0,$ we have
$a_{1} = 3, b_{1} = - 1, c_{1} = - 3 a n d a_{2} = 9, b_{2} = - 3, c_{2} = - 9$
Here, $\frac{a_{1}}{a_{2}} = \frac{3}{9} = \frac{1}{3}$
$\frac{b_{1}}{b_{2}} = \frac{- 1}{- 3} = \frac{1}{3}$
$\frac{c_{1}}{c_{2}} = \frac{- 3}{- 9} = \frac{1}{3}$
$∴$ $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
So, the given equations have infinitely many solutions.
Substitution method
$3 x - y = 3$
$y = 3 x - 3$
on putting the value of $y$ in equation $2$ , we get
$9 x - 3 (3 x - 3) = 9$
$\Rightarrow 9 x - 9 x + 9 = 9$
$\Rightarrow 9 = 9$
Hence, the given equations have infinitely many solutions.

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